Prove: $1+{n\choose 1}\cos\phi+{n\choose 2}\cos2\phi+...+{n\choose n}\cos n\phi=2^n\cos^n\frac{\phi}{2}\cos\frac{n\phi}{2}$

Question

Prove: $\displaystyle 1+{n\choose 1}\cos\phi+{n\choose 2}\cos2\phi+...+{n\choose n}\cos n\phi=2^n\cos^n\frac{\phi}{2}\cos\frac{n\phi}{2}$

I used induction:

For $n=1$ equality holds.

For $n=k\colon$ $$ 1+k\cos\phi+\frac{k(k-1)}{2}\cos2\phi+...+\cos k\phi=2^k \cos^k\frac{\phi}{2}\cos\frac{k\phi}{2} $$

For $n=k+1\colon$ $$(1+k\cos\phi+\frac{k(k-1)}{2}\cos2\phi+...+\cos k\phi)+\cos (k+1)\phi=2^{k+1} \cos^{k+1}\frac{\phi}{2}\cos\frac{(k+1)\phi}{2}$$

This is not correct (example: $k=1,\displaystyle \phi=\frac{\pi}{3}$).

Is it possible to use induction or solving left hand side of equation?

Answer 1

Using Comnplex Number.

Let $$\displaystyle z= e^{i\phi} = \cos \phi+i\sin \phi$$

and $$\Re(z) = \Re(e^{i\phi}) = \cos \phi$$

Now Let $$\displaystyle S = \binom{n}{0}+\binom{n}{1}\Re(e^{i\phi})+\binom{n}{2}\Re(e^{i2\phi})+.......+\binom{n}{n}\Re(e^{in\phi})$$

So we get $$\displaystyle S = \Re[\binom{n}{0}+\binom{n}{1}e^{i\phi}+\binom{n}{2}e^{i2\phi}+......+\binom{n}{n}e^{in\phi}] = \Re[(1+e^{i\phi})^n]$$

Now $$\displaystyle 1+e^{i\phi} = 1+\cos \phi+i\sin \phi = 2\cos^2 \frac{\phi}{2}+i\cdot 2\sin \frac{\phi}{2}\cdot \cos \frac{\phi}{2}$$

So we get $$\displaystyle 1+e^{i\phi} = 2\cos \frac{\phi}{2}[\cos \frac{\phi}{2}+i\sin \frac{\phi}{2}]=2\cos \frac{\phi}{2}\cdot e^{i\frac{\phi}{2}}$$

So We get $$\displaystyle S = \Re[\cos \frac{\phi}{2}\cdot e^{i\frac{\phi}{2}}]^n = 2\cos^n \frac{\phi}{2}\Re[e^{i\frac{n\phi}{2}}] = 2\cos^n \frac{\phi}{2}\left[\cos \frac{\phi}{2}+i\sin \frac{\phi}{2}\right]$$

So we get $$\displaystyle S = 2\cos^n \frac{\phi}{2}\cdot \cos \frac{\phi}{2}$$

Answer 2

$$\sum_{k=0}^n {n \choose k}a^k b^{n-k} = (a+b)^n$$

Your sum is the real part of $$\sum_{k=0}^n {n \choose k} e^{ik \phi} = (1 + e^{i \phi})^n = \left(e^{\frac{1}{2}i \phi}\left(e^{-\frac{1}{2}i \phi} + e^{\frac{1}{2}i \phi}\right)\right)^n$$